$\int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$ is equal to
$\int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$ is equal to
- $x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$
- $x-\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$
- $x+\log \left|\sin \left(x+\frac{\pi}{4}\right)\right|+c$
- $x-\log \left|\sin \left(x+\frac{\pi}{4}\right)\right|+c$
Solution
$\int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$
Let $x+\frac{\pi}{4}=t \Rightarrow x=t-\frac{\pi}{4}$
$
\begin{aligned}
& d x=d t \\
& \int \frac{\sqrt{2} \sin \left(t-\frac{\pi}{4}\right)}{\sin t} d t \\
& \quad=\int \sqrt{2} \cdot \frac{\left(\sin t \cos \frac{\pi}{4}-\cos t \sin \frac{\pi}{4}\right)}{\sin t} d t
\end{aligned}
$
$\begin{aligned} & =\int \sqrt{2} \cdot \frac{\left(\frac{\sin t}{\sqrt{2}}-\frac{\cos t}{\sqrt{2}}\right)}{\sin t} d t \\ & =\int \frac{\sin t-\cos t}{\sin t} d t=\int(1-\cot t) d t \\ & =t-\ln |\sin t|+c \\ & =x+\frac{\pi}{4}-\ln \left|\sin \left(x+\frac{\pi}{4}\right)\right|+c \\ & =x-\log \left|\sin \left(x+\frac{\pi}{4}\right)\right|+c\end{aligned}$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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